Steel wool is placed inside a cylinder filled with molecular oxygen O2, considered as an ideal gas. A piston ensures a constant pressure of the gas. The steel wool reacts with the molecular oxygen to form iron rust Fe2O3, 2Fe +3/2 O2 → Fe2O3 ,The reaction is slow,so that the gas remains at ambient

Question

Steel wool is placed inside a cylinder filled with molecular oxygen [latex]O_2[/latex], considered as an ideal gas. A piston ensures a constant pressure of the gas. The steel wool reacts with the molecular oxygen to form iron rust [latex] Fe_2O_3 [/latex],

[latex] 2 Fe+ \frac{3}{2} O_2 → Fe_2O_3 [/latex].

The reaction is slow, so that the gas remains at ambient temperature [latex]T_0[/latex]. Determine the heat [latex]Q_{if}[/latex], the work [latex]W_{if}[/latex] and the internal energy variation [latex]ΔU_{if}[/latex] in terms of the enthalpy of reaction [latex]ΔH_{if}[/latex] for a reaction involving two moles of iron.

Numerical Application:

[latex] ΔH_{if}[/latex] = −830 kJ, [latex]T_0[/latex] = 25°C.

Accepted Answer

Since the system is coupled to a work reservoir at constant pressure, according to the relation (4.61) the heat transfer [latex]Q_{if}[/latex] is equal to the enthalpy of reaction,

[latex] Q_{if} = ΔH_{if} [/latex] (4.61)

[latex] Q_{if} = ΔH_{if}[/latex] = −830 kJ.

According to relation (2.28), the work [latex]W_{if}[/latex] performed on the ideal gas at constant pressure [latex] p_0[/latex] is given by,

[latex] W_{if} =\int_{i}^{f}{\delta W} = - \int_{V_i}^{V_f}{p(V)dV} [/latex] (reversible adiabatic process) (2.28)

[latex] W_{if} = - \int_{V_i}^{V_f}{p_0 dV} = -P_0 \int_{V_i}^{V_i -ΔV}{dV} = p_0 ΔV [/latex].

When two moles of iron are consumed 3/2 moles of molecular oxygen [latex]O_2[/latex] are oxidised, i.e. ΔN = 3/2. The change in volume of the molecular oxygen ΔV at constant pressure [latex]p_0[/latex] is expressed as,

[latex] p_0 ΔV = Δ N R T_0 [/latex].

Thus, the work [latex]W_{if}[/latex] is recast as,

[latex] W_{if} = p_0 ΔV = Δ N R T_0 [/latex] = 3, 715 kJ.

According to the first law (1.44), the internal energy variation [latex]ΔU_{if}[/latex] is given by,

[latex] ΔU_{if} = W_{if} + Q_{if} [/latex] (closed system) (1.44)

[latex] ΔU_{if} = W_{if} + Q_{if}[/latex] = 2, 885 kJ.

Question 8.6: Steel wool is placed inside a cylinder filled with molecular...

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